What is the integral of 1/(z(z+1)) ?

The integral of 1/(z(z+1)) is ln|z|-ln|z+1|+C

integral of 1/(z(z+1))

\int \frac{1}{z ( z + 1 )} d z

ln ∣ z ∣ - ln ∣ z + 1 ∣ + C

Solution steps

\int \frac{1}{z ( z + 1 )} d z

Take the partial fraction of

\frac{1}{z ( z + 1 )} : \frac{1}{z} - \frac{1}{z + 1}

= \int \frac{1}{z} - \frac{1}{z + 1} d z

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{z} d z - \int \frac{1}{z + 1} d z

\int \frac{1}{z} d z = ln ∣ z ∣

\int \frac{1}{z + 1} d z = ln ∣ z + 1 ∣

= ln ∣ z ∣ - ln ∣ z + 1 ∣

Add a constant to the solution

= ln ∣ z ∣ - ln ∣ z + 1 ∣ + C

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What is the integral of 1/(z(z+1)) ?
The integral of 1/(z(z+1)) is ln|z|-ln|z+1|+C